Total sum except adjacent of a given node in BST

Given a BST and a key Node, find the total sum in BST, except those Node which are adjacent to key Node. 

Examples:

1. First find the total sum of BST 
2. Search the key Node and trace its parent Node. 
3. If the key Node is present then, subtract the sum of its adjacent Node from total sum 
4. If key is not present in BST then return -1.

Implementation:

C++

// C++ program to find total sum except a given Node in BST
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    struct Node *left, *right;
};

// insertion of Node in Tree
Node* getNode(int n)
{
    struct Node* root = new Node;
    root->data = n;
    root->left = NULL;
    root->right = NULL;
    return root;
}

// total sum of bst
int sum(struct Node* root)
{
    if (root == NULL)
        return 0;

    return root->data + sum(root->left) + sum(root->right);
}

// sum of all element except those which are adjacent to key Node
int adjSum(Node* root, int key)
{

    int parent = root->data;

    while (root != NULL) {
        if (key < root->data) {
            parent = root->data;
            root = root->left;
        }
        else if (root->data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjacent Node and
            // subtract from totalSum
            if (root->left != NULL && root->right != NULL)
                return (parent + root->left->data +
                                 root->right->data);

            // if key is leaf
            if (root->left == NULL && root->right == NULL)
                return parent;

            // If only left child is null
            if (root->left == NULL)
                return (parent + root->right->data);

            // If only right child is NULL
            if (root->right == NULL)
                return (parent + root->left->data);
        }

        else {
            parent = root->data;
            root = root->right;
        }
    }

    return 0;
}

int findTotalExceptKey(Node *root, int key)
{
    return sum(root) - adjSum(root, key);
}

// Driver code
int main()
{
    struct Node* root = getNode(15);
    root->left = getNode(13);
    root->left->left = getNode(12);
    root->left->left->left = getNode(11);
    root->left->right = getNode(14);
    root->right = getNode(20);
    root->right->left = getNode(18);
    root->right->right = getNode(24);
    root->right->right->left = getNode(23);
    root->right->right->right = getNode(25);
    int key = 20;
    printf("%d ", findTotalExceptKey(root, key));
    return 0;
}

Java

// Java program to find total sum
// except a given Node in BST
class GFG
{
static class Node
{
    int data;
    Node left, right;
};

// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}

// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;

    return root.data + sum(root.left) +
                       sum(root.right);
}

// sum of all element except those
// which are adjacent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;

    while (root != null)
    {
        if (key < root.data)
        {
            parent = root.data;
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjacent Node and
            // subtract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data +
                                root.right.data);

            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;

            // If only left child is null
            if (root.left == null)
                return (parent + root.right.data);

            // If only right child is null
            if (root.right == null)
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}

static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}

// Driver code
public static void main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    System.out.printf("%d ",
               findTotalExceptKey(root, key));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find total sum
# except a given Node in BST
class getNode:

    def __init__(self, n):

        self.data = n
        self.left = None
        self.right = None

# Total sum of bst
def sum(root):

    if (root == None):
        return 0

    return (root.data + sum(root.left) +
                        sum(root.right))

# Sum of all element except those
# which are adjacent to key Node
def adjSum(root, key):

    parent = root.data

    while (root != None):
        if (key < root.data):
            parent = root.data
            root = root.left
        elif (root.data == key):

            # Key Node matches
            # if the left Node and right Node of key is
            # not None then add all adjacent Node and
            # subtract from totalSum
            if (root.left != None and
               root.right != None):
                return (parent + root.left.data +
                                 root.right.data)

            # If key is leaf
            if (root.left == None and
               root.right == None):
                return parent

            # If only left child is None
            if (root.left == None):
                return (parent + root.right.data)

            # If only right child is None
            if (root.right == None):
                return (parent + root.left.data)

        else:
            parent = root.data
            root = root.right

    return 0

def findTotalExceptKey(root, key):

    return sum(root) - adjSum(root, key)

# Driver code
if __name__ == '__main__':

    root = getNode(15)
    root.left = getNode(13)
    root.left.left = getNode(12)
    root.left.left.left = getNode(11)
    root.left.right = getNode(14)
    root.right = getNode(20)
    root.right.left = getNode(18)
    root.right.right = getNode(24)
    root.right.right.left = getNode(23)
    root.right.right.right = getNode(25)

    key = 20

    print(findTotalExceptKey(root, key))

# This code is contributed by bgangwar59

C#

// C# program to find total sum
// except a given Node in BST
using System;

class GFG
{
class Node
{
    public int data;
    public Node left, right;
};

// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}

// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;

    return root.data + sum(root.left) +
                       sum(root.right);
}

// sum of all element except those
// which are adjacent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;

    while (root != null)
    {
        if (key < root.data)
        {
            parent = root.data;
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjacent Node and
            // subtract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data +
                                 root.right.data);

            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;

            // If only left child is null
            if (root.left == null)
                return (parent + root.right.data);

            // If only right child is null
            if (root.right == null)
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}

static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}

// Driver code
public static void Main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    Console.Write("{0} ",
            findTotalExceptKey(root, key));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to find total sum
// except a given Node in BST
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
};

// Insertion of Node in Tree
function getNode(n)
{
    let root = new Node(n);
    return root;
}

// Total sum of bst
function sum(root)
{
    if (root == null)
        return 0;

    return root.data + sum(root.left) +
                       sum(root.right);
}

// Sum of all element except those
// which are adjacent to key Node
function adjSum(root, key)
{
    let parent = root.data;

    while (root != null)
    {
        if (key < root.data)
        {
            parent = root.data;
            root = root.left;
        }

        // key Node matches
        else if (root.data == key)
        {

            // If the left Node and right Node of key is
            // not null then add all adjacent Node and
            // subtract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data +
                                root.right.data);

            // If key is leaf
            if (root.left == null && root.right == null)
                return parent;

            // If only left child is null
            if (root.left == null)
                return (parent + root.right.data);

            // If only right child is null
            if (root.right == null)
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}

function findTotalExceptKey(root, key)
{
    return sum(root) - adjSum(root, key);
}

// Driver code
let root = getNode(15);
root.left = getNode(13);
root.left.left = getNode(12);
root.left.left.left = getNode(11);
root.left.right = getNode(14);
root.right = getNode(20);
root.right.left = getNode(18);
root.right.right = getNode(24);
root.right.right.left = getNode(23);
root.right.right.right = getNode(25);

let key = 20;
document.write(findTotalExceptKey(root, key));

// This code is contributed by divyeshrabadiya07

</script>

Output

118 

Time Complexity: O(n) + O(h) where n is number of nodes in BST and h is height of BST. We can write time complexity as O(n).